You would like to shoot an orange in a tree with your bow and arrow. The orange

Question

You would like to shoot an orange in a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. On your first try, you fire the arrow at 37.0 m per ssecond at an angle of 30.0 above the horizontal from a height of 1.40 m while standing 41.0 m away. Treating the arrow as a point projectle and neglecting air resistance, what is the height of the arrow once it has traveled the 41.0 m horizontally to the tree? If you fire at the same speed and angle on your second try, how far away could you stand such that the arrow will hit the orange? Assume that the orange remains fixed in place during the arrow's fight. Select all that apply.

Explanation / Answer

Horizontal speed of the arrow, Vx = 37*cos30 = 32.04 m/s.

Vertical speed of the arrow, Vy = 37*sin30 = 18.5 m/s.

Now, time taken by the arrow to travel 41.0 m horizontally:

t = 41.0/32.04 = 1.28 s

Part (A)

Height of the arrow, when it has travelled 41.0 m horizontally

h = 18.5*1.28 - 0.5*9.8*1.28^2 = 15.65 m.

So, the total height of the arrow from ground = 15.65+1.40 = 17.05 m

Part (B)

In order to hit the orange from arrow:

required height = 5.0 - 1.4 = 3.6 m

Now, 3.6 = 18.5t - 4.9t^2

=> 4.9t^2 - 18.5t + 3.6 = 0

=> t = [18.5 + sqrt(342.25 - 70.56)]/9.8 = [18.5+16.48]/9.8 = 3.57 s

second value of t = [18.5-16.48]/9.8 = 0.206 s

Hence distance of the person from orange = 32.04*3.57 or 32.04*0.206

=114.38 m or 6.6 m

So, last two are the correct option.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.